3.203 \(\int \frac {\cos ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=58 \[ -\frac {3 b^3 \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {13}{6};\frac {19}{6};\cos ^2(c+d x)\right )}{13 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{13/3}} \]

[Out]

-3/13*b^3*hypergeom([1/2, 13/6],[19/6],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(13/3)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {16, 3772, 2643} \[ -\frac {3 b^3 \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {13}{6};\frac {19}{6};\cos ^2(c+d x)\right )}{13 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{13/3}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(b*Sec[c + d*x])^(4/3),x]

[Out]

(-3*b^3*Hypergeometric2F1[1/2, 13/6, 19/6, Cos[c + d*x]^2]*Sin[c + d*x])/(13*d*(b*Sec[c + d*x])^(13/3)*Sqrt[Si
n[c + d*x]^2])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx &=b^2 \int \frac {1}{(b \sec (c+d x))^{10/3}} \, dx\\ &=\left (b^2 \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \left (\frac {\cos (c+d x)}{b}\right )^{10/3} \, dx\\ &=-\frac {3 \cos ^5(c+d x) \, _2F_1\left (\frac {1}{2},\frac {13}{6};\frac {19}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{13 b^2 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 60, normalized size = 1.03 \[ -\frac {3 b^2 \sqrt {-\tan ^2(c+d x)} \cot (c+d x) \, _2F_1\left (-\frac {5}{3},\frac {1}{2};-\frac {2}{3};\sec ^2(c+d x)\right )}{10 d (b \sec (c+d x))^{10/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(b*Sec[c + d*x])^(4/3),x]

[Out]

(-3*b^2*Cot[c + d*x]*Hypergeometric2F1[-5/3, 1/2, -2/3, Sec[c + d*x]^2]*Sqrt[-Tan[c + d*x]^2])/(10*d*(b*Sec[c
+ d*x])^(10/3))

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \cos \left (d x + c\right )^{2}}{b^{2} \sec \left (d x + c\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(b*sec(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c))^(2/3)*cos(d*x + c)^2/(b^2*sec(d*x + c)^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(b*sec(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^2/(b*sec(d*x + c))^(4/3), x)

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maple [F]  time = 1.03, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{2}\left (d x +c \right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(b*sec(d*x+c))^(4/3),x)

[Out]

int(cos(d*x+c)^2/(b*sec(d*x+c))^(4/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(b*sec(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^2/(b*sec(d*x + c))^(4/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\cos \left (c+d\,x\right )}^2}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2/(b/cos(c + d*x))^(4/3),x)

[Out]

int(cos(c + d*x)^2/(b/cos(c + d*x))^(4/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(b*sec(d*x+c))**(4/3),x)

[Out]

Integral(cos(c + d*x)**2/(b*sec(c + d*x))**(4/3), x)

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